To solve the system of equations, we can substitute the expression for y in the first equation with the expression for y in the second equation:
x^2 + 4 = 2x + 1
Now we can rearrange the equation by bringing all the terms to one side:
x^2 - 2x + 3 = 0
We can solve this quadratic equation by using the quadratic formula:
x = [-(-2) ± sqrt((-2)^2 - 4(-1)(3))] / 2(-1)
x = [2 ± sqrt(16)] / -2
x = -1 or x = 3
Now we can find the corresponding values of y for each solution:
When x = -1, y = 2(-1) + 1 = -1
When x = 3, y = -3^2 + 4 = -5
Therefore, the solution to the system of equations is (-1, -1) and (3, -5).
To check, we can substitute each solution into the original equations:
When x = -1 and y = -1:
y = -x^2 + 4 becomes -1 = -(-1)^2 + 4, which is true
y = 2x + 1 becomes -1 = 2(-1) + 1, which is true
When x = 3 and y = -5:
y = -x^2 + 4 becomes -5 = -(3)^2 + 4, which is true
y = 2x + 1 becomes -5 = 2(3) + 1, which is true
Therefore, the solutions are valid.
The inverse of a function can be found by switching the roles of x and y and solving for y:
For f(x) = -x^2 + 4, we have:
x = -y^2 + 4
y^2 = 4 - x
y = ± sqrt(4 - x)
So, the inverse of f is f^-1(x) = ± sqrt(4 - x)
For g(x) = x^2 - 6, we have:
x = y^2 - 6
y^2 = x + 6
y = ± sqrt(x + 6)
So, the inverse of g is g^-1(x) = ± sqrt(x + 6)
The domain and range of the original functions and their inverses can follow a pattern, depending on the specific functions. However, in general, the domain and range of a function and its inverse are switched. For example, if the domain of f(x) is all real numbers, then the range of f^-1(x) will be all real numbers. Similarly, if the range of f(x) is all positive numbers, then the domain of f^-1(x) will be all positive numbers.